A233150 Number of ways to write n = k + m (k, m > 0) with 2^k + prime(m) prime.

0, 0, 1, 2, 1, 4, 0, 6, 1, 4, 1, 3, 1, 8, 2, 3, 2, 5, 2, 8, 2, 2, 5, 4, 4, 6, 6, 3, 5, 5, 2, 5, 9, 4, 7, 3, 7, 5, 4, 5, 9, 4, 5, 6, 3, 8, 7, 5, 5, 11, 5, 7, 4, 6, 3, 6, 5, 6, 5, 6, 5, 6, 3, 4, 6, 3, 5, 4, 5, 7, 6, 4, 5, 5, 4, 3, 9, 6, 4, 5, 4, 6, 4, 3, 5, 8, 3, 7, 9, 10, 8, 7, 2, 8, 3, 6, 6, 8, 8, 3

Offset

1,4

Comments

Conjecture: a(n) > 0 except for n = 1, 2, 7.

We have verified this for n up to 3*10^7. For n = 15687374, the least positive integer k with 2^k + prime(n-k) prime is 51299. For n = 28117716, the least positive integer k with 2^k + prime(n-k) prime is 81539.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..6000

Zhi-Wei Sun, On a^n + b*n modulo m, preprint, arXiv:1312.1166 [math.NT], 2013-2014.

Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014-2016.

Example

a(9) = 1 since 9 = 7 + 2 with 2^7 + prime(2) = 128 + 3 = 131 prime.
a(13) = 1 since 13 = 3 + 10 with 2^3 + prime(10) = 8 + 29 = 37 prime.
a(588) = 1 since 588 = 66 + 522 with 2^{66} + prime(522) = 2^{66} + 3739 = 73786976294838210203 prime.
a(1012) = 1 since 1012 = 317 + 695 with 2^{317} + prime(695) = 2^{317} + 5231 prime.

Mathematica

a[n_]:=Sum[If[PrimeQ[2^k+Prime[n-k]], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]

Crossrefs

Cf. A000040, A000079, A231201, A231557, A231577, A233183, A233204, A233206, A233296.

Sequence in context: A336566 A326144 A120112 * A103977 A378600 A109883

Adjacent sequences: A233147 A233148 A233149 * A233151 A233152 A233153

Keyword

nonn

Author

Zhi-Wei Sun, Dec 05 2013

Status

approved